Not really understanding electric potential

[physiccool78]

Homework Statement

The question is attached.

Homework Equations

V = kq/r

The Attempt at a Solution

The analogy to determining the electric potential by thinking of it as lifting a rock (the more distance the higher the potential) makes sense to me, but I am confused by this equation: kq/r. Doesn't this equation imply that the closer the charge is from the source charge, the higher the potential energy? I always had this dilemma and this is an obstacle that's preventing me from understanding better electricity.

I know the answer is B by the way. I just want someone to explain the discrepancy of my understanding and the formula. Thanks![/B]

 

[andrevdh]

Yes the potential gets higher the closer you get, that is you have to do more
work to push a positive test charge closer to the positively charged object.

 

[physiccool78]

Thanks for the reply. So how come the answer to my question is B?

 

[Andrew Mason]

Thanks for the reply. So how come the answer to my question is B?

The field lines represent the magnitude and direction of the electric force on a unit of POSITIVE charge. So, in going from W to Z you can see that work must be done on a positive charge against the field. Therefore the unit of positive charge has a higher potential energy at Z than at W. So the potential (the potential energy per unit charge) of a charge at Z relative to W is positive. This is equivalent to saying that Z is at a higher potential than W OR that the potential difference between Z and W is positive (greater than zero).

AM

 

[physiccool78]

Thanks for the reply. For some reason, my brain thought (because the lines in my picture converged) that the electric field lines actually represented a charge from a negative source, which completely confused me. Thank you so much!

 

[andrevdh]

The field lines start out on a positive charge and terminate on a negative charge,
so that you can think of a positive charge being at the tail ends and negative charge
at the head end of the arrows. So it would require doing positive work being done to
move a positive test charge upstream or against the direction of the arrows. In the
diagram the arrows are electric vectors and not field lines, but they point in the general
direction in which the electric field lines would be. Note the arrows are shorter on the right-
hand side of the diagram than the arrows on the left-hand side, indicating that the electric
field is stronger (magnitude) on the left-hand side of the diagram.

 

[physiccool78]

The field lines start out on a positive charge and terminate on a negative charge,
so that you can think of a positive charge being at the tail ends and negative charge
at the head end of the arrows. So it would require doing positive work being done to
move a positive test charge upstream or against the direction of the arrows. In the
diagram the arrows are electric vectors and not field lines, but they point in the general
direction in which the electric field lines would be. Note the arrows are shorter on the right-
hand side of the diagram than the arrows on the left-hand side, indicating that the electric
field is stronger (magnitude) on the left-hand side of the diagram.

Ok, thanks. So if the electric field is greater at point W, wouldn't that imply that the electric potential is greater as well?
If
E = Vq
Then electric field is directly proportional to potential.

 

[Andrew Mason]

Ok, thanks. So if the electric field is greater at point W, wouldn't that imply that the electric potential is greater as well?
If
E = Vq
Then electric field is directly proportional to potential.

??This is not correct. E = electric force per unit of (positive) charge. V = potential energy per unit charge so Vq is the potential energy of the charge q. Potential is the dot product of force and displacement per unit charge: $V = \int\vec{E}\cdot d\vec{s}$. If the field is uniform, $V_{w-z} = \vec{E}_{w-z}\cdot \vec{s}$ where $\vec{s}$ is the displacement between W and Z.

AM

 

[physiccool78]

??This is not correct. E = electric force per unit of (positive) charge. V = potential energy per unit charge so Vq is the potential energy of the charge q. Potential is the dot product of force and displacement per unit charge: $V = \int\vec{E}\cdot d\vec{s}$. If the field is uniform, $V_{w-z} = \vec{E}_{w-z}\cdot \vec{s}$ where $\vec{s}$ is the displacement between W and Z.

AM

Oh sorry I meant V = Ed.

Anyways the origin to my confusion is this. I will explain with this analogue. mgh = gravitational potential energy. The higher your object, the higher your potential energy. GMm/r^2 = F

GMm/r^2 * r = GMm/r where m is an object and M is the mass of the Earth. For mgh, potential is increased as high increases, but for the other equation, potential is increased when height is decreased. This fact is what confuses me.

 

[andrevdh]

Yes, that is corrrect. The electric potential is taken to zero very far away, that is when r is very large its inverse is very small. Now we approach the positively charged object from afar with a positively charged small test charge, q, and calculate the amount positive work, W, we have to do to bring it up to the point where we want to know what the electric potential is. According to the definition it is

V = W/q

As we approach the positively charge object the repulsion is increasing and we have to do more work and the electric potential increases very fast as we get near it.

 

[Andrew Mason]

Oh sorry I meant V = Ed.

Anyways the origin to my confusion is this. I will explain with this analogue. mgh = gravitational potential energy. The higher your object, the higher your potential energy. GMm/r^2 = F

GMm/r^2 * r = GMm/r where m is an object and M is the mass of the Earth. For mgh, potential is increased as high increases, but for the other equation, potential is increased when height is decreased. This fact is what confuses me.

The difference is that gravitational force between masses is attractive. So as you increase separation between unit masses, potential energy per unit mass (ie. potential) increases (work is required to increase separation). The opposite applies to unit (positive) charges.

AM